Problem: $f(x, y) = (\sin(y), -x\cos(y))$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(\cos(y), -\cos(y) + x\sin(y))$ (Choice B) B $(0, -\cos(y))$ (Choice C) C $(0, \sin(y))$ (Choice D) D $(\cos(y), x\sin(y))$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y) = (f_0(x, y), f_1(x, y)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $y$, we'll treat $x$ as if it were a constant. Therefore, $f_y = (\cos(y), x\sin(y))$.